Let $f(z)=\sum_{n\geq 0}a_n z^n$ be a Taylor series with rational coefficients with infinitely non-zero $a_n$ which converges
in a small neighboorhood around $0$. Furthermore, assume that
\begin{align*}
f(z)=\frac{P(z)}{Q(z)},
\end{align*}
where $P(z)$ and $Q(z)$ are coprime monic complex polynomials. By developing $\frac{P(z)}{Q(z)}$ as a power sereis around $0$ and comparing it with $f(z)$ we obtain infinitely many polynomial equations in the roots of $P(z)$ and $Q(z)$ which are equal to rational numbers so this seems to force the roots of $P(z)$ and $Q(z)$ to be algebraic numbers.

I think what is being asked is to prove that if a sequence of rational numbers satisfies a recurrence with complex coefficients, then it must satisfy a recurrence with rational coefficients.
–
Gjergji ZaimiJan 22 '11 at 18:51

Yes Gjergji, I think that you may rephrase the problem in these terms
–
Hugo ChapdelaineJan 22 '11 at 18:53

2 Answers
2

Let there be two fields $k\subset K$, and let $f\in k[[x]]$ be a formal power series with coefficients in $k$. If $f\in K(x)$ (rational functions with coefficients in $K$) then $f\in k(x)$. A proof of this is given in J.S. Milne's notes on Etale Cohomology (lemma 27.9).

Well, I think there is a simpler argument. For a power series $g(x)\in\mathbb{C}[[x]]$
and $\sigma\in Aut(\mathbb{C})$ (note that except for the complex conjugation or the identity $\sigma$ is not continuous!) we may define define the power series with coefficients twisted by $\sigma$ which we denote by $g^{\sigma}(x)$. Now an element in $Aut(\mathbb{C})$ respect finite sum and products so it follows from that,
that for all $\sigma\in Aut(\mathbb{C})$ one has
$$
f^{\sigma}(z)=\frac{P^{\sigma}(z)}{Q^{\sigma}(z)}.
$$
From this (and the unique factorization of $\mathbf{C}[x]$) it follows that $P(z)$ and $Q(z)$ have rational coefficients.

Yes, and this works in Gjergji's more general setting, too. I would emphasize that $K[[x]]$ is a UFD, it is implicitly used in your proof.
–
GH from MOJan 23 '11 at 12:35

Yes, I did not check it but the proof might carry over to power series ring in many variables.
–
Hugo ChapdelaineJan 23 '11 at 14:54

There is something which bothers me. You use the fact that the subfield of $\C$ fixed by $Aut(\C)$ is $\Q$. Why is it true ? For exemple it is false if you replace $\C$ by $\R$.
–
Auguste Hoang DucJan 27 '11 at 22:06

Yes you are right, I use the fact that the fixed field of $Aut(C)$ is $Q$. The fact that $Aut(R)={Id}$ is not a problem since you may work in a suitable algebraic closure and as you know $C$ is an algebraic closure of $R$. I guess that in general if you have a field $k\subseteq K$ then you want to show the existence of a field $L$ which contains $K$ such that $Aut_k(L)=k$. Once you have that the proof works.
–
Hugo ChapdelaineJan 28 '11 at 0:46

To Hugo: You probably want that $L^{Aut_k(L)}=k$. But such an $L$ only exists if the field extension $K/k$ is separable.
–
ACLNov 15 '12 at 22:55